Transformers wired in series

Suppose I have multiple identical transformers. Each will drop the supply voltage down to the utilization voltage. Each has the same capacity and
impedance.
Now suppose I take 2 (or more) of them and wire all the primary windings in series ... and wire all the secondary windings in series. I fully expect I will have the same voltage ratio.
But here is my real question. Assuming the supply voltage is the same as would work with one transformer, what happens to the fault current level when N transformers are wired in series like this? Assume the supply has theoretically infinite fault current availability to focus examination of the fault current to just this transformer arrangement.
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On 4/19/07 7:56 AM, in article snipped-for-privacy@news1.newsguy.com,

This sure sounds like a homework problem. If this is a real problem, what is the application? Before helping, I would like to know what YOU have done so far.
Bill -- Fermez le Bush--about two years to go.
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| On 4/19/07 7:56 AM, in article snipped-for-privacy@news1.newsguy.com,
| |> Suppose I have multiple identical transformers. Each will drop the supply |> voltage down to the utilization voltage. Each has the same capacity and |> impedance. |> |> Now suppose I take 2 (or more) of them and wire all the primary windings in |> series ... and wire all the secondary windings in series. I fully expect I |> will have the same voltage ratio. |> |> But here is my real question. Assuming the supply voltage is the same as |> would work with one transformer, what happens to the fault current level |> when N transformers are wired in series like this? Assume the supply has |> theoretically infinite fault current availability to focus examination of |> the fault current to just this transformer arrangement. | | This sure sounds like a homework problem. If this is a real problem, what is | the application? Before helping, I would like to know what YOU have done so | far.
It is not a homework problem. It is exploring a theory I was told could not be done, but still suspect that it can be done. What I have "done" is come up with a theory about what this would do. I'll compare my theory to what experts say, if/when they say anything about this. I doubt it is useful for any real applications except for a limited degree of voltage increase.
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----------------------------
wrote:

--------------- If I read what you mean correctly, you are considering the case of N- (V1/V2) transformers with a series connection on both primary and secondary and using the combination as a V1/V2 transformer -providing -in the ideal case no different from a single unit. Since the transformers are not ideal, all you are doing is increasing the effective impedance of the combination with respect to a single transformer (you are not increasing the power rating unless you are applying a higher total voltage which you are not doing in your question) so the result will be poorer voltage regulation (more of a voltage drop under load) and a lower short circuit current compared to a single unit. Total no load current may be lower as flux density in each core will be lower but otherwise nothing gained. Sketch the circuit using the mutual inductance model with primary and secondary electrically isolated rather than the usual T model that is generally used for a transformer. You should see what is going on.
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| If I read what you mean correctly, you are considering the case of N- | (V1/V2) transformers with a series connection on both primary and secondary | and using the combination as a V1/V2 transformer -providing -in the ideal | case no different from a single unit. Since the transformers are not ideal, | all you are doing is increasing the effective impedance of the combination | with respect to a single transformer (you are not increasing the power | rating unless you are applying a higher total voltage which you are not | doing in your question) so the result will be poorer voltage regulation | (more of a voltage drop under load) and a lower short circuit current | compared to a single unit. Total no load current may be lower as flux | density in each core will be lower but otherwise nothing gained. | Sketch the circuit using the mutual inductance model with primary and | secondary electrically isolated rather than the usual T model that is | generally used for a transformer. You should see what is going on.
I am wanting to confirm my belief that it really would have a lower short circuit current. But then what I want to do is see what kind of design would mimic this. I suspect a design intended for a higher voltage, but used at a lower voltage, would be like that. For example, consider a transformer with a 7200 volt primary and 240 volt secondary. If it is supplied with 480 volts, the output should be 16 volts to a light load. If the transformer impedance is 4%, then the short circuit current would be within it's loading rating, right?
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Suppose that you have a 7200/240V 10KVA transformer with 4% impedance. It would have an impedance seen from the primary of 207 ohms (4% of 5184 ohms). It's rated current (primary) would be 1.4A and its short circuit current would be 35A Now, at 480V the short circuit current would be 480/207 =2.32A primary which would be 66%ABOVE its rating (corresponding to (480/7200)*35) meaning that magic smoke won't appear quite as soon as it would at 35A-but it still would occur. The open circuit voltage would be 16V as you indicate. You would have to limit the input voltage to 290V in order to stay within the 1.4A rating. Changing the transformer base voltage doesn't change the actual impedance. if you rerated this transformer to 480V, the rated current would still be 1.4A (unless you want to let magic smoke out-admittedly not as soon as at 35A short circuit current) and the % impedance on the new base would be higher (60%) . If you stacked a bunch in series as you indicated, then the performance would be extremely poor. Is this what you want?
I don't see what you expect to gain. A stack such as this is performance-wise and economically, inferior to a properly designed single unit.
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| Suppose that you have a 7200/240V 10KVA transformer with 4% impedance. | It would have an impedance seen from the primary of 207 ohms (4% of 5184 | ohms). It's rated current (primary) would be 1.4A and its short circuit | current would be 35A | Now, at 480V the short circuit current would be 480/207 =2.32A primary which | would be 66%ABOVE its rating (corresponding to (480/7200)*35) meaning that | magic smoke won't appear quite as soon as it would at 35A-but it still | would occur. The open circuit voltage would be 16V as you indicate. You | would have to limit the input voltage to 290V in order to stay within the | 1.4A rating. | Changing the transformer base voltage doesn't change the actual impedance. | if you rerated this transformer to 480V, the rated current would still be | 1.4A (unless you want to let magic smoke out-admittedly not as soon as at | 35A short circuit current) and the % impedance on the new base would be | higher (60%) .
OK, maybe I have to take this a step further ... 240 volts in, 8 volts out.
| If you stacked a bunch in series as you indicated, then the performance | would be extremely poor. Is this what you want?
What is the difference between one transformer rated 15000 VA, 7200 volts primary, 240 volts secondary, 4% impedance ... and ... 30 transformers rated 500 VA, 240 volts primary, 8 volts secondary, 4% impedance, wired with both primary and secondary in series as I've described (besides the latter being a rats nest).
How many primary winding turns would the former need to have? How many would the latter need to have? Would the latter times 30 equal the former?
One of the things that got me thinking on this was seeing a pole pig near a road, dedicated powing a few lights on a sign. I figured there was not more than a couple KW of light there, if that. The pole pig was rather small. Yet I was trying to imagine what kind of primary winding it would have given that it was probably being supplied with no less than 7200 volts. And I was thinking that, as the KVA rating goes down, number of turns goes up, and at voltages of 7200 and up, that's going to soon reach a point where the size can't get smaller because of the heavier insulation needed with the many turns (though I suppose if very carefully designed, not so much insulation would be needed between adjacent turns).
| I don't see what you expect to gain. A stack such as this is | performance-wise and economically, inferior to a properly designed single | unit.
And it may not be the economical way even for what I have in mind, which is to power a load which will approach being a short circuit, such as an electric arc. But I am only thinking about this in a theoretical sense because I did not believe it when someone told me it was not possible to make a transformer that would convert a voltage, limit the current, and do this within it's load handling capacity, without using an additional pure inductance in series with the secondary. Well, I think it certainly is possible. Maybe what they meant is: it is not ecomically practical.
How much will the power meter spin if the load side terminals are bolted together in a short circuit (before magic smoke gets out)? I would think not much if any because there's no appreciable voltage. Depending on the fault current taking place, it can still dissipate some heat, perhaps quite fast. It would be an interesting race to see who lets the magic smoke out first: the meter, the wiring, or the transformer.
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| As far as your objective of limiting current to the rated current- probably | you should be looking at how welding transformers are designed. They are | rated on a different basis than normal power transformers as they are not | used as normal power transformers. --
I would guess they have a different way of limiting current, such as extra inductance. For things I want to experiment with, that may well be the way to go.
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That works but there are alternatives which would be useful if one is building many units. One way is to design so leakage reactance is relatively high. On open circuit the voltage would be normal but on load, the voltage regulation would be poor- essentially limiting short circuit current. Probably cheaper than a normal transformer as one could skimp on core material.
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On 20 Apr 2007 21:56:37 GMT, snipped-for-privacy@ipal.net wrote:

Wrong...
The definition of transformer impedance:
"The percentage of rated primary voltage that will generate rated current in a short-circuited secondary."
A 4% transformer with a 7200 volt primary will deliver rated current to a shorted secondary when the primary voltage is 288 volts.
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wrote: | On 20 Apr 2007 21:56:37 GMT, snipped-for-privacy@ipal.net wrote: |
|> |>| If I read what you mean correctly, you are considering the case of N- |>| (V1/V2) transformers with a series connection on both primary and secondary |>| and using the combination as a V1/V2 transformer -providing -in the ideal |>| case no different from a single unit. Since the transformers are not ideal, |>| all you are doing is increasing the effective impedance of the combination |>| with respect to a single transformer (you are not increasing the power |>| rating unless you are applying a higher total voltage which you are not |>| doing in your question) so the result will be poorer voltage regulation |>| (more of a voltage drop under load) and a lower short circuit current |>| compared to a single unit. Total no load current may be lower as flux |>| density in each core will be lower but otherwise nothing gained. |>| Sketch the circuit using the mutual inductance model with primary and |>| secondary electrically isolated rather than the usual T model that is |>| generally used for a transformer. You should see what is going on. |> |>I am wanting to confirm my belief that it really would have a lower short |>circuit current. But then what I want to do is see what kind of design |>would mimic this. I suspect a design intended for a higher voltage, but |>used at a lower voltage, would be like that. For example, consider a |>transformer with a 7200 volt primary and 240 volt secondary. If it is |>supplied with 480 volts, the output should be 16 volts to a light load. |>If the transformer impedance is 4%, then the short circuit current would |>be within it's loading rating, right? | | Wrong... | | The definition of transformer impedance: | | "The percentage of rated primary voltage that will generate rated | current in a short-circuited secondary." | | A 4% transformer with a 7200 volt primary will deliver rated current | to a shorted secondary when the primary voltage is 288 volts.
OK, my numbers were a bit off (not sure where I goofed the arithmetic). But I had the concept.
Now I shall build (in my mind) a transformer (array) that will have a maximum fault current of somewhere in the range of 1 to 5 milliamps, using step up instead of step down.
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snipped-for-privacy@ipal.net wrote:

Total primary R increases. Total secondary R increases. Higher voltage drops on both sides, for a given I.
Fault current would depend on where/what the fault was. For example, say the fault is primary shorted to ground. If it happens in xformer 1 (the one connected directly to hot) no difference vs if it was the sole xformer. But if it happens in transformer N, it has all the preceeding primary windings creating a larger voltage drop, therefore lower fault current.
Ed
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| snipped-for-privacy@ipal.net wrote: |> Suppose I have multiple identical transformers. Each will drop the supply |> voltage down to the utilization voltage. Each has the same capacity and |> impedance. |> |> Now suppose I take 2 (or more) of them and wire all the primary windings in |> series ... and wire all the secondary windings in series. I fully expect I |> will have the same voltage ratio. |> |> But here is my real question. Assuming the supply voltage is the same as |> would work with one transformer, what happens to the fault current level |> when N transformers are wired in series like this? Assume the supply has |> theoretically infinite fault current availability to focus examination of |> the fault current to just this transformer arrangement. |> | | Total primary R increases. Total secondary R increases. | Higher voltage drops on both sides, for a given I. | | Fault current would depend on where/what the fault was. | For example, say the fault is primary shorted to ground. | If it happens in xformer 1 (the one connected directly to | hot) no difference vs if it was the sole xformer. But | if it happens in transformer N, it has all the preceeding | primary windings creating a larger voltage drop, therefore | lower fault current.
What if the fault is a bolted short on the secondary terminals right at the transformer(s)? That should be the worst case other than internal faults inside the windings, I would think.
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snipped-for-privacy@ipal.net wrote:

Lower current than if it was the same short against a single xformer because of the increased secondary resistance when the xformers are in series.
Ed
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----------------------------


---------------------------- Bolted short on N transformers - N times the impedance. 1/N times the fault current. That is what ehjsr and I are trying to say. Draw the circuit setup with primaries and secondaries isolated from each other- not as in the common T model. It should be fairly apparent then.
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| Bolted short on N transformers - N times the impedance. 1/N times the fault | current. That is what ehjsr and I are trying to say. Draw the circuit setup | with primaries and secondaries isolated from each other- not as in the | common T model. It should be fairly apparent then.
It is apparent. It seemed that way long before I first posted this. But I wasn't sure if I had overlooked some other concept that would get in the way (besides the economic practicality issue). So I asked. Seems to be nothing unexpected would get in the way.
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in
I
ASSuming the transformer cores don't saturate, the load on each transformer will be "reflected" back to the primary side.
The reflected impedance will be in the ratio of the square of turns ratio.
If the turns ratio is 1/1 then it's as if the loads were put in series. The high resistance/impedance load will see higher voltage and the lower impedance load will see lower voltage. If the voltage drop across the primary exceeds the voltage rating (plus "margin") the transformer will saturate twice each cycle and effectively short out for the last part of each half cycle.
If you want to "solve" the problem, you first solve to non-saturation and then determine the saturation point and continue the solution from there. It's possible the more than one transformer could saturate if you have many in series.
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| The reflected impedance will be in the ratio of the square of turns ratio. | | If the turns ratio is 1/1 then it's as if the loads were put in series. | The high resistance/impedance load will see higher voltage and the lower | impedance load will see lower voltage. If the voltage drop across the | primary exceeds the voltage rating (plus "margin") the transformer will | saturate twice each cycle and effectively short out for the last part of | each half cycle.
The secondaries are also in series, supplying one load.
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